O 3/2/2017 11:55 PM A o/3 Print Calculator Periodic Table Question 1 of 3 Saplin

Question

O 3/2/2017 11:55 PM A o/3 Print Calculator Periodic Table Question 1 of 3 Sapling Learning this question hasbeencustomized by Pamela Mills at city University of New York CUNY, Lehman Coll The reaction of the strong acid HBr with the strong base KOH is: HBr(aq) KOH (aq) Compute the pH of the resulting solution if 75mL of 0.67M acid is mixed with 33mL of base Let's do this steps Number How many moles of acid before reaction? Number How many moles of base before reaction? What is the limiting reactant? Number How many moles of the excess reagent after reaction? Number What is the concentration of the excess reagent after reaction? Number What is the pH of the resulting solution? O Previous Check Answer Give Up & View So Hint Gradebook Map A Next Ext

Explanation / Answer

first find the moles using formula

moles = Molarity x volume in liters

moles of HBr = 0.67 M x 0.075 L = 0.05025 mol

moles of base (KOH) = 0.39 M x 0.033 L = 0.01287 mol

from the balanced equation 1 mol of Acid react with one mle of base so

limiting agent is KOH

excess reagent

0.01287 mol of base react with 0.01287 mol of Acid

remaining acid moles = 0.05025 mol - 0.01287 mol = 0.03738 mol

total volume = 75+33 = 108 mL = 0.108 L

concentration of excess reagent is

M = 0.03738 mol / 0.108 L = 0.3461 M

remaining agent is Acid

pH

-log[H+] = -log[0.3461] = 0.46

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